An elevator can carry a maximum load of 1800 kg (elevator + pasenger) is moving up with a constant speed of `2ms^(-1)`. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power..

Explanation

The downward force on the elevator is F `mg + F_(f) = (1800 xx 10) + 4000 = 22000 N` The motor must supply enough power to balance this force. Hence , P = F.v `= 22000 xx 2 = 44000 W` = 59 hp